MATH1251 Questions HELP (1 Viewer)

leehuan · Aug 20, 2016

Setting up the ODE problem again

$It is estimated that the population growth rate of a certain developing country will fall linearly from 2\% to 1.5\% per year over the next decade. The present population is 10 million. What is the estimated population in 10 years time?$

I have a feeling that the last two sentences are just making it an IVP problem. But yeah based off the first sentence can someone please tell me where they got this from?

$\frac{dy}{dt}=(0.02-0.0005t)y$

InteGrand · Aug 20, 2016

leehuan said:
Setting up the ODE problem again

$It is estimated that the population growth rate of a certain developing country will fall linearly from 2\% to 1.5\% per year over the next decade. The present population is 10 million. What is the estimated population in 10 years time?$

I have a feeling that the last two sentences are just making it an IVP problem. But yeah based off the first sentence can someone please tell me where they got this from?

$\frac{dy}{dt}=(0.02-0.0005t)y$

$\noindent Let $t$ be time measured in years and $y$ the population. The population growth rate in this question is referring to the instantaneous proportional growth rate. If this were a constant $k$, then we would have $y^\prime = ky$ (standard growth/decay ODE when the (instantaneous proportional) growth rate is $k$). But we are told that this $k$ is actually not constant, but rather a (varying) function of $t$, $k(t)$. Since it starts at $2\% = 0.02$, and falls linearly at a rate of $\frac{0.5}{10}\%$ per year $\Big{(}$i.e. $0.05\% = 0.0005$ per year$\Big{)}$, we have $k(t) = 0.02 - 0.0005t$, for $t$ between $0$ and $10$. So the required ODE is $y^\prime = k(t)y = \left(0.02 - 0.0005t\right)y$, for $t$ between $0$ and $10$.$

1008 · Aug 20, 2016

InteGrand said:
It approaches K. leehuan typoed the solution, it should be +y₀ in the denominator. (To see this, note that if we sub. in t = 0, we are supposed to get y = y₀.)

Oh alright, that makes sense then...

1008 · Aug 20, 2016

Got another ODE:

$\\\text{Use the substitution }v= \frac{\mathrm{d} y}{\mathrm{d} t} \text{ to solve}\\ \frac{\mathrm{d^2} x}{\mathrm{d} t^2}+\omega ^2x=0.$

leehuan · Aug 20, 2016

1008 said:
Got another ODE:

$\\\text{Use the substitution }v= \frac{\mathrm{d} y}{\mathrm{d} t} \text{ to solve}\\ \frac{\mathrm{d^2} x}{\mathrm{d} t^2}+\omega ^2x=0.$

Lol even my tutor reckons that question has been typo-d.
_______________________

I was just wondering, is it possible to apply least squares on this problem?

$$The differential equation $\frac{dy}{dx}=\lambda \frac{y}{x}$ is employed as a simple model for comparative growth.$\\ $a) Solve: Answer is $y=Ax^\lambda$

$$b) For the fiddler crab, the weights of the large claw, $y$ grams, and the body (without claw), $x$ grams, were found to be:$\\ \begin{align*}x=58 &\iff y=5\\ x=300 &\iff y=78\\ x=536 &\iff y=196\\ x=1080&\iff y=537\\ x=1449&\iff y=773\\ x=2233&\iff y=1380\end{align*}$

$Show that the above model appears reasonable in this example. Determine approximately $\lambda$ and the constant of integration.$

(I don't mind if no attempt is actually made at this question, with or without lssq)

InteGrand · Aug 20, 2016

1008 said:
Got another ODE:

$\\\text{Use the substitution }v= \frac{\mathrm{d} y}{\mathrm{d} t} \text{ to solve}\\ \frac{\mathrm{d^2} x}{\mathrm{d} t^2}+\omega ^2x=0.$

$\noindent By inspection, the solution is $c_1 \cos \omega t + c_2 \sin \omega t$, if $\omega >0$. (Standard SHM solution.)$

$$\noindent To derive this via the proposed substitution (though it's faster to just use the standard method of solving second-order ODE's with constant coefficients using the characteristic equation etc.; this substitution is basically how a HSC proof of the SHM result would go.), recall that if $v = \dot{y}$, then $\ddot{y} = \frac{\mathrm{d}}{\mathrm{d}y}\left(\frac{1}{2}v^2\right)$ (standard HSC 3U result from motion topic). (Using dots to denote time derivatives.) Then the ODE becomes $\frac{\mathrm{d}}{\mathrm{d}y}\left(\frac{1}{2}v^2\right) = -\omega ^2 y$. So $\frac{1}{2} v^2 = -\omega ^2 \cdot \frac{1}{2}y^2 + \frac{1}{2}C$, for some constant $C$. So $v^2 = C -\omega^2 y^2$. Let's just assume $\omega >0$. Note $C$ must be positive in order for the last equation to hold (except in the trivial solution case where $y$ is identically $0$), so we can just write that arbitrary positive constant as $\omega^2 a^2$ for some $a>0$.$

$\noindent Now, $v^2 = C -\omega^2 y$. So $v = \pm \sqrt{a^2 \omega^2 -\omega^2 y^2} = \pm \omega \sqrt{a^2 - y^2}$. Let's look at the positive root (you can show as an exercise that we get the same solution if we also look at the negative root). Writing $v = \frac{\mathrm{d}y}{\mathrm{d}t}$, this DE becomes $\frac{\mathrm{d}y}{\mathrm{d}t} = \omega \sqrt{a^2 -y^2}$. This is now a separable DE, which you should be able to solve.$

boredofstudiesuser1 · Aug 20, 2016

What year of maths is this?

InteGrand · Aug 20, 2016

leehuan said:
Lol even my tutor reckons that question has been typo-d.
_______________________

I was just wondering, is it possible to apply least squares on this problem?

$$The differential equation $\frac{dy}{dx}=\lambda \frac{y}{x}$ is employed as a simple model for comparative growth.$\\ $a) Solve: Answer is $y=Ax^\lambda$

$$b) For the fiddler crab, the weights of the large claw, $y$ grams, and the body (without claw), $x$ grams, were found to be:$\\ \begin{align*}x=58 &\iff y=5\\ x=300 &\iff y=78\\ x=536 &\iff y=196\\ x=1080&\iff y=537\\ x=1449&\iff y=773\\ x=2233&\iff y=1380\end{align*}$

$Show that the above model appears reasonable in this example. Determine approximately $\lambda$ and the constant of integration.$

(I don't mind if no attempt is actually made at this question, with or without lssq)

$\noindent Yeah, least squares can be used. Least squares just refers to choosing values of parameters ($A$ and $\lambda$ here) so that the \emph{sum of squared (vertical) distances} (with these vertical distances being called \textsl{residuals} in statistics) between the graph with these parameter values and the actual data points is minimised. (Further reading may be done here:$

https://en.wikipedia.org/wiki/Least_squares .)

$\noindent However here, the sought function is not linear in its parameters (it is in $A$, but not in $\lambda$), so this is a \textsl{non-linear least squares} problem. We could linearise it by taking logs so that we are instead fitting $Y = a + \lambda X$, where $Y = \ln y$, $X = \ln x$, and $a$ and $\lambda$ are what we want to fit (so now the equation is linear in these parameters), with $a = \ln A$. Since this is a linear least squares problem, we can find the least-squares solution for $a$ and $\lambda$ using the matrix method you asked about in a thread a few months ago. But I don't think these will correspond to the actual least squares values of $A$ and $\lambda$ in the original function. One way to approach the non-linear problem is to just find the function defining the sum of squared residuals from the given data (as a function of $A$ and $\lambda$, and then trying minimising this two-variable function using multivariable calculus methods.$

Further info on non-linear least squares, including what the 'normal equations' are in the non-linear case: https://en.wikipedia.org/wiki/Non-linear_least_squares .

Linear least-squares: https://en.wikipedia.org/wiki/Linear_least_squares_(mathematics) .

leehuan · Aug 20, 2016

Makes sense

I was thinking along the lines of the residual.
______________

I think I just need more guidance with modelling. Solving ODEs aren't too hard in general.

$An initially unpolluted lake of $10^9$ litres has a river flowing through it at $1,000,000$ litres per day. A factory is built which discharges $10,000$ litres per day of pollutant into the lake. Assuming that the total volume of liquid in the lake remains constant at $10^9$ litres...$

$\frac{dy}{dt}=\overbrace{10^4}^{\text{obvious}}-\underbrace{\frac{y}{10^9}\times 1.01 \times 10^6}_{\text{How does the river give this?}}$

(y is the volume of pollutant after time t days)

edit: resolved by Paradoxica through private conversation

1008 · Aug 20, 2016

boredofstudiesuser1 said:
What year of maths is this?

First year lol (MATH1251 at UNSW)

InteGrand said:
$\noindent By inspection, the solution is $c_1 \cos \omega t + c_2 \sin \omega t$, if $\omega >0$. (Standard SHM solution.)$

$$\noindent To derive this via the proposed substitution (though it's faster to just use the standard method of solving second-order ODE's with constant coefficients using the characteristic equation etc.; this substitution is basically how a HSC proof of the SHM result would go.), recall that if $v = \dot{y}$, then $\ddot{y} = \frac{\mathrm{d}}{\mathrm{d}y}\left(\frac{1}{2}v^2\right)$ (standard HSC 3U result from motion topic). (Using dots to denote time derivatives.) Then the ODE becomes $\frac{\mathrm{d}}{\mathrm{d}y}\left(\frac{1}{2}v^2\right) = -\omega ^2 y$. So $\frac{1}{2} v^2 = -\omega ^2 \cdot \frac{1}{2}y^2 + \frac{1}{2}C$, for some constant $C$. So $v^2 = C -\omega^2 y^2$. Let's just assume $\omega >0$. Note $C$ must be positive in order for the last equation to hold (except in the trivial solution case where $y$ is identically $0$), so we can just write that arbitrary positive constant as $\omega^2 a^2$ for some $a>0$.$

$\noindent Now, $v^2 = C -\omega^2 y$. So $v = \pm \sqrt{a^2 \omega^2 -\omega^2 y^2} = \pm \omega \sqrt{a^2 - y^2}$. Let's look at the positive root (you can show as an exercise that we get the same solution if we also look at the negative root). Writing $v = \frac{\mathrm{d}y}{\mathrm{d}t}$, this DE becomes $\frac{\mathrm{d}y}{\mathrm{d}t} = \omega \sqrt{a^2 -y^2}$. This is now a separable DE, which you should be able to solve.$

InteGrand said:
$\noindent Yeah, least squares can be used. Least squares just refers to choosing values of parameters ($A$ and $\lambda$ here) so that the \emph{sum of squared (vertical) distances} (with these vertical distances being called \textsl{residuals} in statistics) between the graph with these parameter values and the actual data points is minimised. (Further reading may be done here:$

https://en.wikipedia.org/wiki/Least_squares .)

$\noindent However here, the sought function is not linear in its parameters (it is in $A$, but not in $\lambda$), so this is a \textsl{non-linear least squares} problem. We could linearise it by taking logs so that we are instead fitting $Y = a + \lambda X$, where $Y = \ln y$, $X = \ln x$, and $a$ and $\lambda$ are what we want to fit (so now the equation is linear in these parameters), with $A = \ln a$. Since this is a linear least squares problem, we can find the least-squares solution for $a$ and $\lambda$ using the matrix method you asked about in a thread a few months ago. But I don't think these will correspond to the actual least squares values of $A$ and $\lambda$ in the original function. One way to approach the non-linear problem is to just find the function defining the sum of squared residuals from the given data (as a function of $A$ and $\lambda$, and then trying minimising this two-variable function using multivariable calculus methods.$

Further info on non-linear least squares, including what the 'normal equations' are in the non-linear case: https://en.wikipedia.org/wiki/Non-linear_least_squares .

Linear least-squares: https://en.wikipedia.org/wiki/Linear_least_squares_(mathematics) .

Thanks for that. Yeah, it's probably just easier to not use the substitution and just solve it like a second order ODE.

boredofstudiesuser1 · Aug 20, 2016

Wow... I'm screwed... what degree would you have to do this in?

leehuan · Aug 20, 2016

Also to the people actually in the course, I'm just helping advertise the PASS classes that they are running. They are worth going imo though.

This question seems pretty tedious to write out (as in, both the qn and ans). I was gonna work my way up by finding m₀=f(t) but then I realised that if I did that then I'd be solving an endless stream of first order linear ODEs without guaranteeing I get a pattern. Is there a way to force the pattern out?

RenegadeMx · Aug 20, 2016

leehuan said:
Also to the people actually in the course, I'm just helping advertise the PASS classes that they are running. They are worth going imo though.

This question seems pretty tedious to write out (as in, both the qn and ans). I was gonna work my way up by finding m₀=f(t) but then I realised that if I did that then I'd be solving an endless stream of first order linear ODEs without guaranteeing I get a pattern. Is there a way to force the pattern out?

trying to get on the exec team eh?

1008 · Aug 20, 2016

I've got a couple of ODEs too.

1)Use the substitution to solve:

$xyy''=yy'+x(y')^2;u=y'/y \text{ or }u=\ln y$

$\\\text{2) Suppose that }\alpha(x)\text{ is a solution to the equation }\\ \frac{\mathrm{d^2} u}{\mathrm{d} x^2} +b(x) \frac{\mathrm{d} u}{\mathrm{d} x} +c(x)u=0\\\text{(a) Use the substitution }u(x)=\alpha(x)v(x)\text{ to write down a first order equation for } \frac{\mathrm{d} v}{\mathrm{d} x}\\\text{What method can you see to solve this equation?}\\\text{(b) Verify that }u(x)=x^3\text{ is one of the solutions to the differential equation }\\x^2\frac{\mathrm{d^2} u}{\mathrm{d} x^2}-4x\frac{\mathrm{d} u}{\mathrm{d} x}+6u=0\\\text{Hence use the technique of (a) to find the general solution.}$

leehuan · Aug 20, 2016

1008 said:
I've got another ODE too. Use the substitution to solve:

$xyy''=yy'+x(y')^2;u=y'/y \text{ or }u=\ln y$

I couldn't find this question anywhere in the course pack. Which question is it?

$\text{ Let }y=e^{ u } \implies \frac{dy}{dx}=e^u\frac{du}{dx}\implies \frac{d^2y}{dx^2}=e^u\left(\frac{dy}{dx}+\frac{d^2y}{dx^2}\right)\\ \text{ Then }x\frac { du }{ dx } +x\frac { d^{ 2 }u }{ dx^{ 2 } } =\frac { du }{ dx } +\left( \frac { du }{ dx } \right) ^{ 2 }$

$\text{Let }v=\frac{du}{dx}\implies \frac{dv}{dx}=\frac{d^2u}{dx^2}\\ xv+x\frac{dv}{dx}=v+xv^2\\ v+\frac{dv}{dx}=\frac{v}{x}+xv^2$

That is not exact. And it's definitely not linear. But I don't see how it's separable.

$\text{Let }v=jx\implies \frac{dv}{dx}=x\frac{dj}{dx}+j\\ j + \frac{dj}{dx}=x^2j^2$

And I've lost my motivation to continue.

Edit: Wait

$\text{Let }j=\frac{c}{x} \implies \frac{dj}{dx}=-\frac{c}{x^2}\frac{dc}{dx}\\ \frac{c}{x}-\frac{c}{x^2}\frac{dc}{dx}=c^2$

This technically is first order linear, but my last two substitutions do no effect overall. So I'm inclined to think I might've made a mistake

RenegadeMx said:
trying to get on the exec team eh?

Maybe

Paradoxica · Aug 20, 2016

1008 said:
I've got a couple of ODEs too.

1)Use the substitution to solve:

$xyy''=yy'+x(y')^2;u=y'/y \text{ or }u=\ln y$

$\\\text{2) Suppose that }\alpha(x)\text{ is a solution to the equation }\\ \frac{\mathrm{d^2} u}{\mathrm{d} x^2} +b(x) \frac{\mathrm{d} u}{\mathrm{d} x} +c(x)u=0\\\text{(a) Use the substitution }u(x)=\alpha(x)v(x)\text{ to write down a first order equation for } \frac{\mathrm{d} v}{\mathrm{d} x}\\\text{What method can you see to solve this equation?}\\\text{(b) Verify that }u(x)=x^3\text{ is one of the solutions to the differential equation }\\x^2\frac{\mathrm{d^2} u}{\mathrm{d} x^2}-4x\frac{\mathrm{d} u}{\mathrm{d} x}+6u=0\\\text{Hence use the technique of (a) to find the general solution.}$

I didn't use the substitution but that's just moi....

$$\noindent$ xyy'' = yy' + x(y')^2 $ (divide both sides by $xyy'$)$ \\\\ \frac{y''}{y'} = \frac{1}{x} + \frac{y'}{y} \\\\ $Integrate both sides with respect to $x$ to obtain:$ \\\\ \log{y'} + \mathcal{C} = \log{x} + \log{y} \\\\ ky' = xy \\\\ \frac{y'}{y} = k_1 x \\\\ \log{y} + \mathcal{C}_1 = k_2 x^2 \\\\ Ay = e^{bx^2}$

1008 · Aug 20, 2016

1008 said:
I've got a couple of ODEs too.

1)Use the substitution to solve:

$xyy''=yy'+x(y')^2;u=y'/y \text{ or }u=\ln y$

$\\\text{2) Suppose that }\alpha(x)\text{ is a solution to the equation }\\ \frac{\mathrm{d^2} u}{\mathrm{d} x^2} +b(x) \frac{\mathrm{d} u}{\mathrm{d} x} +c(x)u=0\\\text{(a) Use the substitution }u(x)=\alpha(x)v(x)\text{ to write down a first order equation for } \frac{\mathrm{d} v}{\mathrm{d} x}\\\text{What method can you see to solve this equation?}\\\text{(b) Verify that }u(x)=x^3\text{ is one of the solutions to the differential equation }\\x^2\frac{\mathrm{d^2} u}{\mathrm{d} x^2}-4x\frac{\mathrm{d} u}{\mathrm{d} x}+6u=0\\\text{Hence use the technique of (a) to find the general solution.}$

leehuan said:
I couldn't find this question anywhere in the course pack. Which question is it?

$\text{ Let }y=e^{ u } \implies \frac{dy}{dx}=e^u\frac{du}{dx}\implies \frac{d^2y}{dx^2}=e^u\left(\frac{dy}{dx}+\frac{d^2y}{dx^2}\right)\\ \text{ Then }x\frac { du }{ dx } +x\frac { d^{ 2 }u }{ dx^{ 2 } } =\frac { du }{ dx } +\left( \frac { du }{ dx } \right) ^{ 2 }$

$\text{Let }v=\frac{du}{dx}\implies \frac{dv}{dx}=\frac{d^2u}{dx^2}\\ xv+x\frac{dv}{dx}=v+xv^2\\ v+\frac{dv}{dx}=\frac{v}{x}+xv^2$

That is not exact. And it's definitely not linear. But I don't see how it's separable.

$\text{Let }v=jx\implies \frac{dv}{dx}=x\frac{dj}{dx}+j\\ j + \frac{dj}{dx}=x^2j^2$

And I've lost my motivation to continue.

Edit: Wait

$\text{Let }j=\frac{c}{x} \implies \frac{dj}{dx}=-\frac{c}{x^2}\frac{dc}{dx}\\ \frac{c}{x}-\frac{c}{x^2}\frac{dc}{dx}=c^2$

This technically is first order linear, but my last two substitutions do no effect overall. So I'm inclined to think I might've made a mistake

Maybe

Yeah it's a weird question...I tried both the substitutions suggested but failed to get to a solution.

Did you have a look at this one? I added it later on...

1008 said:
$\\\text{2) Suppose that }\alpha(x)\text{ is a solution to the equation }\\ \frac{\mathrm{d^2} u}{\mathrm{d} x^2} +b(x) \frac{\mathrm{d} u}{\mathrm{d} x} +c(x)u=0\\\text{(a) Use the substitution }u(x)=\alpha(x)v(x)\text{ to write down a first order equation for } \frac{\mathrm{d} v}{\mathrm{d} x}\\\text{What method can you see to solve this equation?}\\\text{(b) Verify that }u(x)=x^3\text{ is one of the solutions to the differential equation }\\x^2\frac{\mathrm{d^2} u}{\mathrm{d} x^2}-4x\frac{\mathrm{d} u}{\mathrm{d} x}+6u=0\\\text{Hence use the technique of (a) to find the general solution.}$

1008 · Aug 20, 2016

Paradoxica said:
I didn't use the substitution but that's just moi....

$$\noindent$ xyy'' = yy' + x(y')^2 $ (divide both sides by $xyy'$)$ \\\\ \frac{y''}{y'} = \frac{1}{x} + \frac{y'}{y} \\ $Integrate both sides with respect to $x$ to obtain:$ \\\\ \log{y'} + \mathcal{C} = \log{x} + \log{y} \\\\ ky' = xy \\ \frac{y'}{y} = k_1 x \\ \log{y} + \mathcal{C}_1 = k_2 x^2$

Woah.

Wait: if we integrate both sides wrt x, don't we treat y as a constant?

Paradoxica · Aug 20, 2016

1008 said:
Woah.

Wait: if we integrate both sides wrt x, don't we treat y as a constant?

No. y is a function of x... Hence, thereby, the thing follows...

1008 · Aug 20, 2016

Paradoxica said:
No. y is a function of x... Hence, thereby, the thing follows...

Yeah thanks for clarifying that. By doing waay too many exact ODEs (treating y as constant) I completely forgot about that fact.
Any thoughts on the other question though?

MATH1251 Questions HELP (1 Viewer)

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